(10%) Problem 7: Consider a parallel plate capacitor having plates of area 0.000

Question

(10%) Problem 7: Consider a parallel plate capacitor having plates of area 0.000485 m2 that are separated by 0.085 mm of Teflon The dielectric constant of Teflon is 2.1 Find the capacitance in F Grade Summary Deductions Potential 0% 100% Submissions cosO cotanO asin0 acosO atanacotan) sinh( coshO tanhO cotanh0 Degrees O Radians sin tanO Attempts remaining: 3 (4% per attempt) detailed view 1 2 3 0 END DELI CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 2 Feedback: 3%-deduction per feedback.

Explanation / Answer

Capacitacne of a capacitor is given by:

C = k*e0*A/d

k = dielectric constant of material = 2.1

e0 = 8.85*10^-12

A = Area = 0.000485 m^2

d = space between plates = 0.085 mm = 0.085*10^-3 m

So,

C = 2.1*8.85*10^-12*0.000485/(0.085*10^-3)

C = 1.06*10^-10 F

C = 1.06*10-10 F = 1.06e-10 F

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